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3 years ago

Factorize

?}^{2} - x - 6" alt="x {?}^{2} - x - 6" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

dmitriy555 [2]3 years ago

5 0

<h2><u>Given</u> :-</h2>

x^2 - x - 6

<h2><u>Solution</u> :-</h2>

Factorisation :-

x^2 - x - 6

= (x - 3)(x + 2)

$\\$

Answered by - ItzMaster

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First one is d and second one is c

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3 years ago

Consider the following data set. The data were actually collected in pairs, and each row represents a pair.

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Answer:

Step-by-step explanation:

Hello!

Using the data sets you have to analyze them as if they are two independent samples.

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Assuming that both data sets are from a normal distribution and both population variances, although unknown, are equal, the statistic to use is:

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The p-value for the two-tailed test is:

P( $t_{18}$ ≤-0.875) + P( $t_{18}$ ≥0.875)= P( $t_{18}$ ≤-0.875) + (1 - P( $t_{18}$ ≤0.875))= 0.1965 + ( 1 - 0.8035)= 0.393

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The p-value for the two-tailed test is:

P( $t_{9}$ ≤-2.05) + P( $t_{9}$ ≥2.05)= P( $t_{9}$ ≤-2.05) + (1 - P( $t_{9}$ ≤2.05))= 0.0353 + (1 - 0.9647)= 0.0706

Comparing the p-value with the significance level, the hypothesis test is not significant. Meaning that the population mean of the difference between group 1 and group 2 is equal to cero. There is no difference between the two groups.

The value of the statistic in "a" is greater than the value of the statistic obtained in "b". Since there are two samples used an "a" the degrees of freedom of the test are the double as the ones used in "b". The p-value obtained in "b" is around half of the p-value of "a".

The main difference is that in "a" you compared two different samples but in "b" you compared two dependent samples when analyzing paired data the "effect of the individual" is removed from the equation.

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